Infinte Area but finite Volume

[drcrabs]

Whence evaluating the area under the curve

y=\frac{1}{x} \\\ \mbox{for} \\\ 1 \leq x < \infty

it evaluates to \infty

But when evaluating the volume using

Volume = \pi \int y^2 dx \\\ \mbox{on} \\\ a \leq x < b

hence


Volume = \pi \int \frac{1}{x^2} \\dx

hence

Volume = \pi [-\ \frac{1}{x}] \\\ \mbox{on} \\\ 1 \leq x < \infty

hence

Volume = \pi [0 - - 1] = \pi

A finite value!

Im having trouble comprehending such concepts and ideas.
Can someone please explain?

[Brad Barker]

Whence evaluating the area under the curve

y=\frac{1}{x} \\\ \mbox{for} \\\ 1 \leq x < \infty

it evaluates to \infty

But when evaluating the volume using

Volume = \pi \int y^2 dx \\\ \mbox{on} \\\ a \leq x < b

hence


Volume = \pi \int \frac{1}{x^2} \\dx

hence

Volume = \pi [-\ \frac{1}{x}] \\\ \mbox{on} \\\ 1 \leq x < \infty

hence

Volume = \pi [0 - - 1] = \pi

A finite value!

Im having trouble comprehending such concepts and ideas.
Can someone please explain?


for a more direct "blow your mind" property, the surface area of that solid of revolution is infinite (and the volume, like you said, finite).

things dealing with infinity get pretty strange.

there are fractals that exhibit similar properties, like infinite surface area but zero volume and such.

i deal with it by just casting it off as math. :biggrin: